# Why does in string theory the amount of supersymmetry have to be $\cal{N} \leq 2$?

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Why is this, that in string theory the maximum amount of supersymmetry is $\cal{N} = 2$, whereas in supergravity one can have up to $\cal{N} = 8$ ?

asked Apr 16, 2013
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I think its important to know which dimension you are in, knowing only say $\mathcal N=2$ is not enough to determine how much supersymmetry there is and therefore hard to compare to $\mathcal N=8$.

This post imported from StackExchange Physics at 2014-03-09 16:21 (UCT), posted by SE-user Heidar
@Heidar I have read the statement that there can not be more than N eq 2 in string theory as being generally true (it was not further explained). This made me wonder why this is, since supergravity can have (dependent on the dimension?) can have a higher amount of supersymmetry.

This post imported from StackExchange Physics at 2014-03-09 16:21 (UCT), posted by SE-user Dilaton
My point was that those two statements are not necessarily inconsistent with each other, one has to specify the space-time dimension. The fact that $\mathcal N\leq 2$ (understood appropriately) comes from analyzing the spectrum of superstring theory (see here for the results en.wikipedia.org/wiki/…). However, this is a statement in 10 dimensions. If $\mathcal N=1$ that means you have a spinor $Q_\alpha$, but the number of components $\alpha$ is given by representation theory of Clifford algebra. In higher dimensions, spinors have (cont.)

This post imported from StackExchange Physics at 2014-03-09 16:21 (UCT), posted by SE-user Heidar
more components. As you see in the above link, type I string theory has 16 supercharges ($\mathcal N=1$) and type II has 32 ($\mathcal N=2$), this is related to spinors in 10 dimensions (which I think have 16 components/spinor dimension). In lower dimensions you can still have, say, 32 supercharges but you can't do it with only one spinor since the spinors have fewer components there, thus you must have $\mathcal N\geq 2$. When people talk about $\mathcal N=8$ SUGRA, it's in $D=4$ I think. In four dimensions Dirac spinors have 4 components, so $8\times 4 = 32$ super charges. (cont.)

This post imported from StackExchange Physics at 2014-03-09 16:21 (UCT), posted by SE-user Heidar
The same as $\mathcal N=2$ in D=10. All I'm saying here is correct morally, but there might be some details which aren't completely correct. However, more details can be found in most SUSY notes.

This post imported from StackExchange Physics at 2014-03-09 16:21 (UCT), posted by SE-user Heidar
Correction: "can't do it with only one spinor" should say "can't do it with only two spinors".

This post imported from StackExchange Physics at 2014-03-09 16:21 (UCT), posted by SE-user Heidar
Thanks for these helpful explanation @Heidar.

This post imported from StackExchange Physics at 2014-03-09 16:21 (UCT), posted by SE-user Dilaton

## 1 Answer

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For higher numbers of supersymmetries on the sigma model's world sheet, the target space dimension becomes negative. (Proof is a boring computation.) It's not particularly clear how to interpret this -- e.g., should the theory have a supergravity limit? -- so these theories are typically discarded from the classifications.

This post imported from StackExchange Physics at 2014-03-09 16:21 (UCT), posted by SE-user user1504
answered Apr 16, 2013 by (1,110 points)

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